Problem: Solve the equation. $\dfrac{11}6 =n+\dfrac79$ $n=$
Answer: Let's subtract to get $n$ by itself. $\begin{aligned}\dfrac{11}6& =n+\dfrac79 \\ \\ \dfrac{11}6 {-\dfrac79}&= n+\dfrac79{-\dfrac79}~~~~~{\text{subtract }\dfrac79} \text{ from each side to get } n \text{ by itself }\\ \\ \dfrac{11}6 {-\dfrac79}&= n+\cancel{\dfrac79}{-}\cancel{{\dfrac79}} \\\\ \dfrac{11}6 {-\dfrac79} &= n\end{aligned}$ $\begin{aligned} \dfrac{11}6-\dfrac79 &=\dfrac{11\times3}{6\times3}-\dfrac{7\times2}{9\times2}\\\\ &= \dfrac{33}{18}-\dfrac{14}{18} \\\\ &= \dfrac{19}{18} \end{aligned}$ The answer: $n={\dfrac{19}{18}}$ Let's check to make sure. $\begin{aligned} \dfrac{11}6 &=n+\dfrac79\\\\ \dfrac{11}6&\stackrel{?}{=} {\dfrac{19}{18}}+\dfrac79 \\\\ \dfrac{11}6&\stackrel{?}{=} {\dfrac{19}{18}}+\dfrac{14}{18} \\\\ \dfrac{11}6&\stackrel{?}{=} \dfrac{33}{18} \\\\ \dfrac{11}6 &= \dfrac{11}6 ~~~~~~~~~~~~~~~~\text{Yes!} \end{aligned}$